Subject: Chemistry — Chemical Thermodynamics
Question: One mole of an ideal gas at 900 K, undergoes two
reversible processes, I followed by II, as shown below. If the work done
by the gas in the two processes are same the value of
$\ln\frac{v_{3}}{v_{2}}$ is \_\_\_\_\_\_\_.
[IMAGE] (U: internal energy, S: entropy, p: pressure, V: volume, R: gas
constant)
(Given: molar heat capacity at constant volume, $C_{V,m}$ of the gas
is $\frac{5}{2}R$)
Answer: (10)
Solution: $\mathrm{\Delta}U_{1} = nC_{v}\mathrm{\Delta}T$
$$- 1800\ R = 1 \times \frac{5R}{2} \times \mathrm{\Delta}T$$
$$\mathrm{\Delta}T_{1} = - 720k$$
$$T_{2} - T_{1} = - 720\ \left\lbrack T_{1} = 900k \right\rbrack$$
$T_{2} = 180\ k$.
Process II is reversible isothermal
$$W_{II} = - nRT\ln\left( \frac{V_{3}}{V_{2}} \right)$$
$$W_{II} = W_{II} = - 1800R$$
So $- 1800R = - 180R\ln\left( \frac{V_{3}}{V_{2}} \right)$
So $\ln\left( \frac{V_{3}}{V_{2}} \right) = 10$